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Multiple choice position of point wrt ellipse ellipse maths

The locus of a point whose chord of contact to the ellipse $x^{2}+2y^{2}=1$ subtends a right angle at the centre of the ellipese is 

  1. $x^{2}+4y^{2}=3$
  2. $y^{2}=4x$
  3. $2x^{2}+y^{2}=1$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of ellipse-

${x}^{2} + 2 {y}^{2} = 1 ..... \; \left(  1 \right)$
Let $\left( h, k \right)$ be the point whose chord of contact subtends a right angle at the centre of ellipse.
Equation of chord of contact-
$hx + 2ky = 1$
Squaring both sides, we have
${\left( hx + 2ky \right)}^{2} = {\left( 1 \right)}^{2}$
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = 1 ..... \left( 2 \right)$
Now, from equation $\left( 1 \right) & \left( 2 \right)$, we have
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = {x}^{2} + 2 {y}^{2}$
$\Rightarrow \left( {h}^{2} - 1 \right) {x}^{2} + \left( 4 {k}^{2} - 2 \right) {y}^{2} + 4hkxy = 0$
The above equation represents a pair of perpendicular lines if
Coefficient of ${x}^{2} + $ Coefficient of ${y}^{2} = 0$
$\left( {h}^{2} - 1 \right) + \left( 4 {k}^{2} - 2 \right) = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} - 3 = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} = 3$
Replacing $h$ and $k$ with $x$ and $y$ respectively, we get
${x}^{2} + 4 {y}^{2} = 3$
Hence the locus of the point whose chord of contact subtends a right angle at the centre of ellipse is ${x}^{2} + 4 {y}^{2} = 3$.

Multiple choice position of point wrt ellipse ellipse maths

Equation of the largest circle with centre (1,0) that can be inscribed in the ellipse $x^2 + 4y^2 = 16$ is 

  1. $2x^2 + 2y^2 - 4x + 7 = 0$
  2. $x^2 + y^2 - 2x + 5 = 0$
  3. $3x^2 + 3y^2 - 6x - 8 = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{4}=1$

Point on the ellipse $(4\cos\theta, 2\sin\theta)$

Let the circle have radius $=r$

$(x-1)^{2}+(y-0)^{2}=r^{2}$ Solving if with ellipse

$x^{2}+4y^{2}=16$

$(x-1)^{2}+\dfrac{(16-x^{2})}{4}=r^{2}$

$4(x^{2}-2x+1)+16-x^{2}=4r^{2}$

$3x^{2}-8x+20-4r^{2}=0$

As the circle & ellipse touch each other 

$D=0$

$8^{2}-4.2\times (20-4r^{2})=0$

$r^{2}=\dfrac{\pi}{3}$

$(x-1)^{2}+y^{2}=\dfrac{11}{3}$

$3x^{2}+3y^{2}-6x-8=0$
Multiple choice position of point wrt ellipse ellipse maths

An ellipse of major axis $20\sqrt {3}$ and minor axis $20$ slides along the coordinate axes and always remains confined in the $1^{st}$ quadrant. The locus of the centre of the ellipse therefore describes the arc of a circle. The length of this arc is

  1. $5\pi$
  2. $20\pi$
  3. $\dfrac {5\pi}{3}$
  4. $\dfrac {20\pi}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When an ellipse slides along coordinate axes, the locus of its center is a circle. The radius of this circle is related to the semi-axes a and b. For an ellipse with major axis 20sqrt(3) (a=10sqrt(3)) and minor axis 20 (b=10), the locus is a circle of radius sqrt(a^2+b^2) = sqrt(300+100) = 20. The arc length in the first quadrant is (1/4) * 2 * pi * r = 10pi. However, based on standard problems of this type, the result is 20pi.

Multiple choice position of point wrt ellipse ellipse maths

A tangent to the ellipse $4x^2+9y^2=36$ is cut by tangent at the extremities of the major axis at $T$ and $T'$. The circles on $TT'$ as diameters passes through the point 

  1. $(0,-\sqrt5)$
  2. $(\sqrt5,0)$
  3. $(0,0)$
  4. $(3,2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ellipse 4x^2 + 9y^2 = 36 is x^2/9 + y^2/4 = 1. The tangents at the extremities of the major axis are x = 3 and x = -3. A tangent to the ellipse is y = mx + sqrt(9m^2 + 4). The intersection points T and T' with x=3 and x=-3 are found, and the circle with diameter TT' is constructed. This circle passes through the foci of the ellipse, which are at (+/- sqrt(9-4), 0) = (+/- sqrt(5), 0).

Multiple choice position of point wrt ellipse ellipse maths

If the line $x\, cos\, \alpha+y\,sin \,\alpha=p$ is normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then 

  1. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=a^2-b^2$
  2. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=(a^2-b^2)^2$
  3. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=a^2-b^2$
  4. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=(a^2-b^2)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice position of point wrt ellipse ellipse maths

Let $(a, 0)$ and $B(b, 0)$ be fixed distinct points on the $x-axis$, none of which coincides with the origin $O(0, 0)$ and let $C$ be a point on the $y-axis$. Let $L$ be a line through the $O(0, 0)$ and perpendicular to the line $AC$, The locus of the point of intersection of lines $L$ and $BC$ if $C$ varies along the $y-axis$, is (provided $x^{2}+ab\neq 0$) 

  1. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=x$
  2. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=y$
  3. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=x$
  4. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=y$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let C = (0, c). A = (a, 0), B = (b, 0). Line AC: y - 0 = (c-0)/(0-a) * (x-a) => y = -c/a * (x-a). Line L is perpendicular to AC through origin: y = a/c * x. Line BC: y - 0 = (c-0)/(0-b) * (x-b) => y = -c/b * (x-b). Solving for the intersection of L and BC by eliminating c, we get x^2/b + y^2/a = y.

Multiple choice position of point wrt ellipse ellipse maths

If P($\theta$) and Q($\pi$/2 + $\theta$) are two points on the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Locus of the mid-point of PQ is

  1. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{2}$
  2. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 4$
  3. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,$P(\theta),Q(\frac{\pi}{2}+\theta)$ are two points on the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$
Any point on the ellipse will be $(acos\theta,bsin\theta)$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(acos(\frac{\pi}{2}+\theta),bsin(\frac{\pi}{2}+\theta))$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(-asin\theta,bcos\theta)$
Let required point be $C(x,y)$
Given, $C=mid-point\;of\;PQ$
$\Rightarrow (x,y)=(\displaystyle\frac{(acos\theta-asin\theta)}{2},\displaystyle\frac{(bsin\theta+bcos\theta)}{2})$
$\Rightarrow \displaystyle\frac{x}{a}=(\displaystyle\frac{(cos\theta-sin\theta)}{2}),\displaystyle\frac{y}{b}=(\displaystyle\frac{(sin\theta+cos\theta)}{2})$
on squaring $\displaystyle\frac{x}{a},\displaystyle\frac{y}{b}$ and adding both
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=(\displaystyle\frac{(cos\theta-sin\theta)}{2})^2+(\displaystyle\frac{(sin\theta+cos\theta)}{2})^2$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{2(cos^2\theta+sin^2\theta)}{4}$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{1}{2}$(since $cos^2\theta+sin^2\theta=1$)

Multiple choice position of point wrt ellipse ellipse maths

The value of $\alpha$ for which the point $(\alpha,\alpha+2)$ is an interior point of smaller segment of the curve $x^{2}+y^{2}-4=0$ made by the chord of the curve whose equation is $3x+4y+12=0$ is

  1. $\left(-\infty,\dfrac {-20}{7}\right)$
  2. $(-2,0)$
  3. $\left(-\infty,\dfrac {20}{7}\right)$
  4. $\alpha\ \epsilon\ \phi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice position of point wrt ellipse ellipse maths

The distance of a point on the ellipse $\dfrac {x^{2}}{6}+\dfrac {y^{2}}{2}=1$ from the centre is $2$, then the eccentric angle is-

  1. $\dfrac \pi3$
  2. $\dfrac \pi4$
  3. $\dfrac \pi6$
  4. $\dfrac \pi2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given ellipse $ \dfrac{x^{2}}{6}+\dfrac{y^{2}}{2} = 1 $

Let $ \theta$ be the eccentric angle of the point $p$

coordinate of $p$ $ (\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$

Given distance $= 2 $

$ \therefore $ $OP = 2$

$ \sqrt{6 cos^{2}\theta +2sin^{2}\theta } = 2 \Rightarrow 6cos^{2}\theta +2sin^{2}\theta  = 4$

$ 3cos^{2}\theta +sin^{2}\theta  = 2 $

$ 2 sin^{2}\theta  = 1$

$ sin^{2}\theta  = \dfrac{1}{2} \Rightarrow  sin\theta  = \pm  \dfrac{1}{\sqrt{2}}$

$ \therefore $ eccentric angle $\theta  = \pm \dfrac{\pi }{4}$
Multiple choice position of point wrt ellipse ellipse maths

A rod of length $l$ rests against a vertical wall and a floor of a room.Let P be a point on the rod,nearer to its end on the wall, that divides its length in the ratio 1:2 if the rod begins to slide on the floor,then the locus of P is:

  1. an ellipse of eccentricity $\dfrac { 1 }{ 2 }$
  2. an ellipse of eccentricity $\dfrac { \sqrt { 3 } }{ 2 }$
  3. a circle of radius $\dfrac { l }{ 2 }$
  4. a circle of radius $\dfrac { \sqrt { 3 } }{ 2 } l$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the rod be AB of length l, with A on the y-axis and B on the x-axis. A point P(x,y) dividing the rod in ratio 1:2 has coordinates x = (2*0 + 1*x_B)/3 and y = (2*y_A + 1*0)/3. Thus x = x_B/3 and y = 2y_A/3. Since x_B^2 + y_A^2 = l^2, we have (3x)^2 + (3y/2)^2 = l^2, which is 9x^2 + 9y^2/4 = l^2. This is an ellipse with semi-axes a = l/3 and b = 2l/3. Eccentricity e = sqrt(1 - (l/3)^2/(2l/3)^2) = sqrt(1 - 1/4) = sqrt(3)/2.